Cantor Confusion
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From: Dik T. Winter on 25 Jan 2007 21:06 In article <1169741815.990021.289480(a)v45g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 25 Jan., 17:03, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > Yes? The elements of S as written above are the a_ik for i and k natural > > numbers. > > Look closer: > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > The elements are sequences S_k like a_1k, a_2k, a_3k, ... Yes, I look closer. The elements are digits like a_1k, a_2k, etc. What are the elements of: { 2^k, 3^k, 5^k, 7^k, 11^k, ... | k in N} ? I would say the elements are numbers, that is all numbers that are powers of primes. By what magical notation change do you make it that the elements are sequences? > > > However, if L is a set of > > > limits L_k > > > L = { L_k | k in N } > > > and S is the set of corresponding sequences > > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > > then L cannot be larger than S, because there cannot be more limits > > > than sequences. > > this makes no sense if S is *not* a set of sequences. > > S is a set of sequences S_k = a_1k, a_2k, a_3k, ... You use your own fantasy notation? > > Consider the > > following: > > L = { L_k | k in {0, 1, 2}} > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in {0, 1, 2} } > > > If S is a set the cardinality of S is at most 2. > > S contains all paths in the union of all finite trees T(oo). Pray try first to properly define your set S. As it is your notation is bad. > > If S is an ordered > > *multi-set*, the ordinality is omega. The cardinality of L is 3. > > S is certainly not an ordered set because it contains multiple identical > > elements. > > S is a set of sequences S_k = a_1k, a_2k, a_3k, ... Darn. When I tried to give the definition of S as you require you said I was wrong. Basic set theory: > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } does *not* define S as a set of sequences. Notations like: S = {[a_1k, a_2k, a_3k, ...]| a in {0, 1}, k in N } might do. On the left of the vertical bar are listed the elements, possibly separated by commas, on the right are listed the possible conditions, etc. With your notation, the elements of S are a_ik with i in N (assumed by the ... notation), k in N and a in {0, 1}. So in your notation S has 0 as element when any of the a_ik is 0, and it has 1 as element when any of the a_ik is 1. So the cardinality is at most 2. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: davidmarcus on 25 Jan 2007 22:46 On Jan 25, 7:28 pm, davidmar...(a)alum.mit.edu wrote: > I'm not sure what you mean by "permutation". Do you mean enumeration or > listing? Maybe you mean bijection (a map that is one to one and onto). > Let's be more precise: > > Let's suppose that by "permutation" you mean a bijection . Suppose p is > a permutation of [0,1] and f: N -> R is a surjection. Let g be defined > by g(x) = p(f(x)). Then g is a surjection N -> R. > On the other hand, if h: N -> R is not a surjection, why should this > imply that no surjection exists? Please replace "R" with "[0,1]" each place it appears.
From: mueckenh on 26 Jan 2007 04:24 On 24 Jan., 21:16, Franziska Neugebauer <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > > > 1) Every complete infinite binary tree T (containing all nodes and > > edges) contains all paths. > No we cannot say so. Please define "contains" first. In my definition, a tree has the special structure which I will not repeat as you know it. Therefore the tree T(n) is well-defined by its nodes alone. If you wish we could use different symbols for the tree and for its set of nodes, but I do not consider this necessary or useful (cp. the case omega and aleph_0). As I defined only trees with paths of equal length, we distinguish trees not by single nodes but by subsets of the set of nodes which are called levels, denoted by L(n). Definition of a level is obvious. Definition 1: Tree T(n) is contained in tree T(m) if the set of nodes T(n) is a subset of the set of nodes T(m), i.e., if n =< m. This definition is sufficient to include every m,n in N. Notice: It does not include N. I do not cosider N a natural number. But every level of the tree is mapped by a natural number, hence for every level L(n) we necessarily have n =/= N. As a tree T is a set of nodes and a path P is a set of nodes, the definition of "contains" with respect to trees and paths is now obvious: P c T. > So you commit to David's trees (std-trees + tree-union). No. I do not read the writings of such writers. I commit to my definition, again outlined above. > We cannot say so. *In* std-trees there no paths. The set of paths is > induced by the tree. The trees which I use, uniquely define their paths, which are in these trees. > > > 4) The set of paths in T(oo) is a subset of the countable set of > > finite sets of all paths in the finite trees. > No. The set of paths of T(oo)/G is the full set of paths. Counterexample: Paths with length 1 are not contained in any tree T(n) with n > 1. > > ==> The set of all path is countable. (==> The real numbers are > > countable.) > Non sequitur. Equivocation fallacy. Are you really not able to understand that the set of real numbers is proven countable if the set of real numbers in some interval is proven countable? > > > Going on, we can say: > > > 6) T(oo) = T contains only finite paths. > No! Please give a clear statement: Would you say that the union of all initial segments of paths with length n is a finite path? Yes or no? If yes, then you should prove your claim by giving an upper bound as is usual in finity proofs of set theory (where the only alternatives are finity and actual infinity). If no, then you should try to explain how an infinite number could creep in and can exist unrecognized as a finite natural number among the natural numbers. >The set of paths of T(oo)/G is complete, i.e. contains really all > paths. That is not a contradiction to be introduced by "no". N contains all natural numbers. Nevertheless all natural numbers are finite. T(oo) contains all initial segments of paths with length n, where n in N. Regards, WM
From: mueckenh on 26 Jan 2007 04:29 On 25 Jan., 15:34, Franziska Neugebauer <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > >> >> You have misunderstood the induction principle. It is not made for > >> >> "counting over to the infinite". > > >> > It is valid for all existing natural numbers. Counting over to the > >> > infinite is nonsense.[indentation corrected] Counting occurs in the > >> > finite. > > > What you do _is_ "counting over to the infinite" > > > By no means! Every n in N is finite. > The set of finite trees V* = { T(n) | n e N } is not a finite set. > Induction over its members does not prove a property of V*. > This is quite the same as in the case of the infinite set > N = { n | n e N }. Induction over n for some property p(n) does not > prove p(N). Maybe you forgot that. > I am not interested in any property of what you call the set N. I am interested in a property due to each and every natural number. Believe it or not, it is sufficient to show by induction the finiteness of every natural number in order to prove that every natural number is finite. Whether you believe that the set N has other properties is your personal opinion. It is irrelevant for the consideraton of trees. In particular we can prove by induction that the tree, formally called T(oo), contains all paths if it contains all levels enumerated by natural numbers. We have no use for your set N in the tree (in case it should differ significantly from the ensmble of all natural numbers.) Regards, WM
From: mueckenh on 26 Jan 2007 04:35
On 25 Jan., 15:57, Franziska Neugebauer <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > > If T(oo) as defined by me is T as defined by me, concerning the set of > > nodes and edges, >There is no room in equality for "concerning" this or that. Either to > symbols refer to the same entity or they do not. That is not so obvious. There can be equality with respect to one property but inequality with respect to another. Think of a red circle and a green circle. Two traffic lights may have identical shape but different colour. And just this colour can have great significance. (As a member of an "applied university", as you like to denote it, I am used to give practical examples). But with respect to trees I support your claim. If the shape (the organigram) of the tree is clear and the set of nodes is defined, then the set of paths is uniquely determined, i.e., well defined too. > > > then they are identical with respect to paths too. > > There is no further parameter to fix. If you do not accept this > > identity, then further discussion is not meaningful. >E-qui-vo-ca-tion. Melody? > > > Then you may maintain ZFC and may say that it is free of > > contradictions,You have not shown any yet. > > > but you must tolerate that a unique structure may > > yield different results, i.e. is not unique.I refuse to tolerate that nonsense. Is that last sentence of yours? It is shown here on the screen as if it was a statement of mine, and in fact it expresses my opinion exactly. So I am not sure wether you or I wrote it. > > > My goal is reached. > Daydreaming? > What do you think about Virgil's claim concerning different sets o paths in identical trees? Regards, WM |